Question

Giải pt 

 \(1+\dfrac{2}{x-2}=\dfrac{10}{x+3}-\dfrac{50}{\left(2-x\right)\left(x+3\right)}\)

 \(\dfrac{x^2-3x+5}{x^2-4}=-1\)

Answer 1

a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)

\(\Leftrightarrow x^2+3x-12-10x-30=0\)

\(\Leftrightarrow x^2-7x-42=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};1\right\}\)